9 THE PLANCHEREL-GODEMENT THEOREM 383 or But, by virtue of (15.9.4.2), (V- ng(x)\ng(yj) = 0 for all x9 y in A. Since A/U0 is dense in Hg , it follows that V= 0 and hence F = 0. Q.E.D. Examples (15.9.5) We have seen (15.6.2.4) that a positive linear form /on an algebra with involution A having a unit element always satisfies the condition (B). We recall that if A is in addition a Banach algebra, then the corresponding Hilbert form g also satisfies (U) (15.6.11). If A is a separable Banach algebra with unit element e, then the prehilbert space A/ng and the star algebra jtfg are also separable. The first assertion follows from the continuity of/ (15.6.11), which implies that \\ng(x)\\2 =/(x*jc) £ ||/|| • \\x*x\\ £ \\f\\ • ||x||2 and shows there- fore that the image under ng of a denumerable dense set in A is dense in the prehilbert space A/ng . The separability of jtfg follows from the fact that the representation Ug is continuous (15.5.7) and therefore transforms a denumer- able dense set in A into a denumerable dense set in jfg . Hence the Bochner- Godement theorem can be applied to every positive linear form on A. (15.9.6) If s\-> U(s) is a representation of A on a Hilbert space H, then for each x0 E H the form fXQ(s) = (U(s) • x0 \XQ) satisfies the condition (B), because by (6.2.4) \fxo(s)\2 g ||E/(s) • x0f • IK!]2 = \\x0[\2fxo(s*s). By (15.5.6) this shows that knowledge of the hermitian characters of a commutative algebra A with involution which has a unit element determines all the representations of A. In addition, we have seen in the course of the proof of (15.9.4) that, if/ is a positive linear form satisfying the conditions of (15.9.4(ii)), then the corresponding representation x\-+Ug(x) admits a totalizer a (15.9.4.2). Remarks (15.9.7) It can happen that the formula (15.9.3) (and the conditions of (15.9.2)) are satisfied but that the measure mg is not bounded (and therefore /does not satisfy (B)) (Problem 2). On the other hand, there are examples in which the conditions (U) and (N) are satisfied but (B) is not (Problem 4), and examples in which (B) and (N) are satisfied but (U) is not (Problem 5).ly on S^ to G, and hence,